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我在数组中有一组纬度和经度坐标。当在 Google 地图中绘制为多边形上的点时，它们大致形成一个从西向东指向的不规则椭圆。

我想以最左（西）轴附近的一个点作为原点将多边形旋转到任意度数。

对我来说重要的是保留了总距离 - 应保留以英里为单位的多边形的总长度和宽度，即使多边形的大小（以像素为单位）由于地图投影而明显不同。

我花了很多时间在这个网站上搜索和搜索这个问题的答案，但一直找不到。严格来说，这不是谷歌地图的问题——它只是一个与地图投影上的旋转多边形相关的几何问题。

到目前为止，或多或少是我一直在玩的东西（为了简单起见，有些东西被删除/重命名）：

function transpose_poly() {
//the polygon I'm transposing; these are actually in lng,lat not lat,lng
var poly = [
    [165.2467094000077,11.90511591102683],[165.1960646350092,11.8776472999036],[165.163749733589,11.84385698549516],[165.1260503052001,11.79105209544025],[165.1174590975139,11.73721064669002],[165.1239723221977,11.68443896599666],[165.1455712453083,11.62322111902237],[165.1628352566873,11.60212814234246],[165.2801792409835,11.57112282455308],[165.3469838984795,11.55321856612457],[165.4267372080734,11.49929306693537],[165.565122175408,11.43334434041074],[165.7036021721537,11.37198352732909],[165.7982543390455,11.32757670668951],[165.942446703552,11.2660302522167],[166.0435044916007,11.23207374453692],[166.1628753311194,11.19161490039798],[166.3468035041342,11.13537294275959],[166.432786256031,11.11077339082378],[166.6460225244011,11.09603091173615],[166.9054486129032,11.10353634871533],[167.0953310801652,11.11920326808891],[167.2738338244123,11.14546202299651],[167.5299835821322,11.20017441185735],[167.7587090824888,11.25007287877568],[168.0532186132958,11.32987818697488],[168.3030699093596,11.40339603540862],[168.592055474493,11.49329084618948],[168.8894586866613,11.59767488596071],[169.1097084341002,11.70426500697907],[169.3388671138959,11.8464629880637],[169.47335151263,11.96284699062962],[169.4987805640997,12.00051052731504],[169.5113979458664,12.04997756596092],[169.496674063518,12.07975001861134],[169.4439862794831,12.10473302818016],[169.3792705121883,12.11718325976015],[169.2053586392944,12.12366910168141],[169.0210976722354,12.12171866909852],[168.7390558752391,12.08703266811138],[168.4733370821476,12.04764814638675],[168.1055698159765,12.00021651042535],[167.8745488025422,11.97152786285725],[167.5955303201492,11.94724207538445],[167.1571321566584,11.94152529858467],[166.8673995936747,11.95771709621411],[166.6698153277294,11.98911065050636],[166.4432968316392,12.03361885637251],[166.2604579582592,12.0693923391982],[166.0834945953367,12.09610600014998],[165.9206278637858,12.10688793842689],[165.7421018997999,12.1070118835482],[165.6244752224984,12.09489163127243],[165.4939687494391,12.0642833194958],[165.4055155587918,12.02521842289156],[165.3386147079753,11.97769336095395],[165.278157772496,11.92168821285675],[165.2467094000077,11.90511591102683],
    ];

    var pos = marker.getPosition(); //where it transposes it to
    var marker_lat = pos.lat(); 
    var marker_lng = pos.lng();
    var angle = document.getElementById("poly_angle").value; //rotational angle

    var original_lat = 11.697222; //the original lat/lng of the polygon coordinates above
    var original_lng = 165.27194399999996; //used to move the general polygon to a new set of coords as an offset

    var new_poly = []; //the transformed polygon

//iterate over polygon array, create new array of Google Maps LatLng objects that are transposed and rotated
    for(var i=0; i<poly.length; i++) {
        new_poly.push( rotateLatLng( (poly[i][1]-original_lat)+marker_lat,(poly[i][0].lng()-original_lng)+marker_lng,angle));

    }

    // plot the polygon on the map
    poly_obj = new google.maps.Polygon({
        paths: poly_new,
        strokeColor: "#ff763b",
        strokeOpacity: 1,
        strokeWeight: 1,
        fillColor: "#ff763b",
        fillOpacity: 0.25,
        map: map,
        visible: true
    });

}

//this rotating function is cobbled together from code I found... it doesn't really work. It somewhat rotates it but distorts it terribly and gives really bizarre results
function rotateLatLng (pointLat,pointLng,angle) {
    var pos = marker.getPosition();
    var theX = pointLat;
    var theY = pointLng;
    var rotationTheta = angle;
    var rotationThetaRad = rotationTheta*(Math.PI/180);
    var rotationOriginX = pos.lat();
    var rotationOriginY = pos.lng();

    var newX;
    var newY;

    if (rotationOriginX == 0 && rotationOriginY == 0) {
        newX = theX * Math.cos(rotationThetaRad) - Math.sin(rotationThetaRad) * theY;
        newY = theX * Math.sin(rotationThetaRad) + Math.cos(rotationThetaRad) * theY;
    } else {
        newX = (theX - rotationOriginX) * Math.cos(rotationThetaRad) - (theY - rotationOriginY) * Math.sin(rotationTheta) + rotationOriginX;
        newY = (theX - rotationOriginX) * Math.sin(rotationThetaRad) + (theY - rotationOriginY) * Math.cos(rotationTheta) + rotationOriginY;
    }


  return new google.maps.LatLng(newX,newY);
}

我认为上述方法根本不一定是正确的方法。任何指针都会有所帮助。

这个问题的一个更简化的版本是说，如果我有一个 lat1,lng1 的原点和 lat2,lng2 的另一个目标点，我如何计算 lat3,lng3 被定义为 lat1 之间的相同距离,lng1 和 lat2,lng2 但在任意角度？因为如果我知道如何做到这一点，那么将其应用于整个多边形应该很容易。

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